Analytical Chemistry

How many milliliters of 0.202 M KOH should be added to 500.0 mL of 0.0233 M
tartaric acid (C4H6O6; FW 150.087) to adjust the pH to 2.75? Ka1 = 1.0 x 10-13 and Ka2 = 4.6 x10-5.

For this one, the Ka1 seems sort of off. The pka would be 13 which doesn't make sense.

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asked by Micki
  1. Tartaric acid is a dibasic acid (two carboxyl COOH groups). You want to add enough acid to completely neutralize the first H and leave the second. That formula is too long to write; let's call it H2C. We add KOH to neutralize the first H to make KHC. Then we make the buffer out of the KHC and C^2-
    mL x M = mL x M
    500 x 0.0233 = mL x 0.202 which gives you about 11.65 (but you can do it more accurately and watch the significant figures). Then we make the buffer from that.
    ........HC^- + OH^- ==> C^2- + H2O
    I.......11.65...0.......0........0
    add.............x.................
    C........-x....-x.......x.........x
    E.......11.65-x..0.......x........x

    Plug all of this into the HH equation and solve for x.
    2.75 = pK2 + log(base)/(acid)
    base = x
    acid = 11.65-x
    x = mmols KOH. Convert to mL of the 0.202M stuff and ADD to the amount needed for the first equivalence point.
    Then I would work backwards, plug in the numbers, and see if it really does produce a pH of 2.75.

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    posted by DrBob222
  2. I ended up with KOH mL being about 1.5mL. The equiv. point would be around 57mL. I see the answer is to be around 20mL. I have no idea how this comes about.

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    posted by Micki

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